Lorentz force in Stern-Gerlach experiment from electrons and nucleus to atom In wikipedia, it is said

The experiment is normally conducted using electrically neutral particles such as silver atoms. This avoids the large deflection in the path of a charged particle moving through a magnetic field and allows spin-dependent effects to dominate.[7][8]

But if I tried to add the Lorentz force from electrons and nuclei to the atom, it looks confusing. For example, a hydrogen atom
$$
\mathbf{F}_e = q_e \mathbf{E} + q_e \mathbf{v}_e \times \mathbf{B} + \mathbf{F}_{p\,\, to  \,\,  e}
$$
$$
\mathbf{F}_p = q_p \mathbf{E} + q_p \mathbf{v}_p  \times \mathbf{B} + \mathbf{F}_{e\,\, to  \,\,  p}
$$
$e$ and $p$ stand for electron and proton. It is known $q_e = - q_p$ and I can expect $\mathbf{F}_{p\,\, to  \,\,  e} = - \mathbf{F}_{e\,\, to  \,\,  p}$. But, it is unlikely $ \mathbf{v}_e =   \mathbf{v}_p $.
I can think in terms of the Ehrenfest theorem to make the above argument somehow more quantum. But, this is also unlikely to reconcile this issue. So the question is, how to obtain the net Lorentz force for a neutral atom to be zero.
 A: The electrons in an atom do not have a velocity in the classical sense. They exist in a superposition of velocities and the expectation value of the velocity, i.e. their average velocity, is the same as the velocity of the atom. This also applies to the nucleus.
You say:

it is unlikely $\mathbf v_e = −\mathbf v_p$

but if we replace $\mathbf v_e$ and $\mathbf v_p$ by their expectation values they will be the same and hence the average Lorentz force on the protons and electrons will be equal and opposite.
A: The argument is not that total Lorentz force on neutral atom is zero, but that by atom being neutral, this total Lorentz force is much smaller than it would be if the atom carried net charge, and second, that it is so small that it is negligible when compared to force due to magnetic field on intrinsic magnetic moment.
The first part is easy - the force is smaller than for a charged system for obvious reason - oppositely charged particles have all roughly the same laboratory "macroscopic" velocity  $\mathbf V$ in the experiment, so those parts of all the Lorentz forces on all particles inside that are proportional to $\mathbf V$ cancel each other out.
However, there is possibly a non-zero force remainder due to the fact that particles' velocities can differ somewhat.
It then becomes the question of whether this force remainder is negligible when compared to force due to intrinsic magnetic moment (spin) in magnetic field gradient. This is much harder question to analyze exactly in quantum theory and it is not easy to answer off-hand, since expectation value of the Lorentz magnetic force is not, in general, the product of $\langle \Delta \mathbf v \rangle$ and $\langle \mathbf B(\mathbf r)\rangle$:
$$
e\bigg\langle (\mathbf v_p - \mathbf v_e) \times \mathbf B(\mathbf r) \bigg\rangle \neq e \langle \mathbf v_p - \mathbf v_e\rangle \times \langle\mathbf B(\mathbf r) \rangle.
$$
However, we can make a classical estimate. Net Lorentz magnetic force due to differing velocities can be reformulated as force on orbital magnetic moment implied by those different velocities, in gradient of magnetic field, described by the well-known formula
$$
\mathbf F_{B~gradient~on~\mu_{orb}} = \boldsymbol{\mu}_{orb} \cdot \nabla \mathbf B.
$$
The "spin force" is the same thing, but due to intrinsic magnetic moment $\boldsymbol{\mu}_{spin}$. So we can say, in classical theory, that non-spin forces affecting the atom trajectory (really due to Lorentz magnetic forces not cancelling each other perfectly) are certainly negligible when orbital magnetic moment is negligible to spin magnetic moment.
Borrowing facts from quantum theory, this seems to be possible in some cases, e.g. when the atom is in ground state with zero orbital angular momentum in direction of magnetic field gradient, but non-zero intrinsic angular momentum in that same direction. But this is not necessarily true for all atoms in all states. It seems that for the silver atoms in the SG experiment, it is always assumed that their angular momentum component in direction of the field gradient is zero, so the only important contribution to force that deflects the atoms is due to spin magnetic moment.
